Precalculus Examples

$\log_{2} (3) \cdot \log_{3} (4) \cdot \log_{4} (5) \cdot \log_{5} (6) \cdot \log_{6} (7)$

Step 1

Multiply $\log_{2} (3) \log_{3} (4) \log_{4} (5) \log_{5} (6)$ by $\log_{6} (7)$ .

$\log_{2} (3) \log_{3} (4) \log_{4} (5) \log_{5} (6) \log_{6} (7)$

Step 2

Rewrite $\log_{2} (3)$ using the change of base formula.

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Step 2.1

The change of base rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$ , and $x$ is greater than $0$ .

$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} \log_{3} (4) \log_{4} (5) \log_{5} (6) \log_{6} (7)$

Step 2.2

Substitute in values for the variables in the change of base formula, using $b = 10$ .

$\frac{\log (3)}{\log (2)} \log_{3} (4) \log_{4} (5) \log_{5} (6) \log_{6} (7)$

Step 3

Rewrite $\log_{3} (4)$ using the change of base formula.

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Step 3.1

The change of base rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$ , and $x$ is greater than $0$ .

$\frac{\log (3)}{\log (2)} \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} \log_{4} (5) \log_{5} (6) \log_{6} (7)$

Step 3.2

Substitute in values for the variables in the change of base formula, using $b = 10$ .

$\frac{\log (3)}{\log (2)} \cdot \frac{\log (4)}{\log (3)} \log_{4} (5) \log_{5} (6) \log_{6} (7)$

Step 4

Cancel the common factor of $\log (3)$ .

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Step 4.1

Cancel the common factor.

$\frac{\log (3)}{\log (2)} \cdot \frac{\log (4)}{\log (3)} \log_{4} (5) \log_{5} (6) \log_{6} (7)$

Step 4.2

Rewrite the expression.

$\frac{1}{\log (2)} \log (4) \log_{4} (5) \log_{5} (6) \log_{6} (7)$

Step 5

Combine $\frac{1}{\log (2)}$ and $\log (4)$ .

$\frac{\log (4)}{\log (2)} \log_{4} (5) \log_{5} (6) \log_{6} (7)$

Step 6

Rewrite $\log (4)$ as $\log (2^{2})$ .

$\frac{\log (2^{2})}{\log (2)} \log_{4} (5) \log_{5} (6) \log_{6} (7)$

Step 7

Expand $\log (2^{2})$ by moving $2$ outside the logarithm.

$\frac{2 \log (2)}{\log (2)} \log_{4} (5) \log_{5} (6) \log_{6} (7)$

Step 8

Cancel the common factor of $\log (2)$ .

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Step 8.1

Cancel the common factor.

$\frac{2 \log (2)}{\log (2)} \log_{4} (5) \log_{5} (6) \log_{6} (7)$

Step 8.2

Divide $2$ by $1$ .

$2 \log_{4} (5) \log_{5} (6) \log_{6} (7)$

Step 9

Rewrite $\log_{4} (5)$ using the change of base formula.

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Step 9.1

The change of base rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$ , and $x$ is greater than $0$ .

$2 \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} \log_{5} (6) \log_{6} (7)$

Step 9.2

Substitute in values for the variables in the change of base formula, using $b = 10$ .

$2 \frac{\log (5)}{\log (4)} \log_{5} (6) \log_{6} (7)$

Step 10

Multiply $2 \frac{\log (5)}{\log (4)}$ .

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Step 10.1

Combine $2$ and $\frac{\log (5)}{\log (4)}$ .

$\frac{2 \log (5)}{\log (4)} \log_{5} (6) \log_{6} (7)$

Step 10.2

Simplify $2 \log (5)$ by moving $2$ inside the logarithm.

$\frac{\log (5^{2})}{\log (4)} \log_{5} (6) \log_{6} (7)$

Step 11

Raise $5$ to the power of $2$ .

$\frac{\log (25)}{\log (4)} \log_{5} (6) \log_{6} (7)$

Step 12

Rewrite $\log (25)$ as $\log (5^{2})$ .

$\frac{\log (5^{2})}{\log (4)} \log_{5} (6) \log_{6} (7)$

Step 13

Expand $\log (5^{2})$ by moving $2$ outside the logarithm.

$\frac{2 \log (5)}{\log (4)} \log_{5} (6) \log_{6} (7)$

Step 14

Rewrite $\log (4)$ as $\log (2^{2})$ .

$\frac{2 \log (5)}{\log (2^{2})} \log_{5} (6) \log_{6} (7)$

Step 15

Expand $\log (2^{2})$ by moving $2$ outside the logarithm.

$\frac{2 \log (5)}{2 \log (2)} \log_{5} (6) \log_{6} (7)$

Step 16

Cancel the common factor of $2$ .

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Step 16.1

Cancel the common factor.

$\frac{2 \log (5)}{2 \log (2)} \log_{5} (6) \log_{6} (7)$

Step 16.2

Rewrite the expression.

$\frac{\log (5)}{\log (2)} \log_{5} (6) \log_{6} (7)$

Step 17

Rewrite $\log_{5} (6)$ using the change of base formula.

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Step 17.1

The change of base rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$ , and $x$ is greater than $0$ .

$\frac{\log (5)}{\log (2)} \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} \log_{6} (7)$

Step 17.2

Substitute in values for the variables in the change of base formula, using $b = 10$ .

$\frac{\log (5)}{\log (2)} \cdot \frac{\log (6)}{\log (5)} \log_{6} (7)$

Step 18

Cancel the common factor of $\log (5)$ .

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Step 18.1

Cancel the common factor.

$\frac{\log (5)}{\log (2)} \cdot \frac{\log (6)}{\log (5)} \log_{6} (7)$

Step 18.2

Rewrite the expression.

$\frac{1}{\log (2)} \log (6) \log_{6} (7)$

Step 19

Combine $\frac{1}{\log (2)}$ and $\log (6)$ .

$\frac{\log (6)}{\log (2)} \log_{6} (7)$

Step 20

Rewrite $\log_{6} (7)$ using the change of base formula.

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Step 20.1

The change of base rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$ , and $x$ is greater than $0$ .

$\frac{\log (6)}{\log (2)} \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}$

Step 20.2

Substitute in values for the variables in the change of base formula, using $b = 10$ .

$\frac{\log (6)}{\log (2)} \cdot \frac{\log (7)}{\log (6)}$

Step 21

Cancel the common factor of $\log (6)$ .

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Step 21.1

Cancel the common factor.

$\frac{\log (6)}{\log (2)} \cdot \frac{\log (7)}{\log (6)}$

Step 21.2

Rewrite the expression.

$\frac{1}{\log (2)} \log (7)$

Step 22

Combine $\frac{1}{\log (2)}$ and $\log (7)$ .

$\frac{\log (7)}{\log (2)}$

Step 23

The result can be shown in multiple forms.

Exact Form:

$\frac{\log (7)}{\log (2)}$

Decimal Form:

$2.80735492 \dots$